- #1

- 1,030

- 4

How do you show that the surreal numbers form a proper class?

- Thread starter Dragonfall
- Start date

- #1

- 1,030

- 4

How do you show that the surreal numbers form a proper class?

- #2

matt grime

Science Advisor

Homework Helper

- 9,395

- 3

Since there are 'more' ordinals than cardinals, and there is a proper class of cardinals (if there is a set of cardinals, C, what is the cardinality of the power set of C?), this becomes easy - if I've understood the one thing I read about surreal numbers and some link to ordinals, and that;s a big if.

- #3

- 1,030

- 4

- #4

- 355

- 3

Never mind, I was reading the Wiki article, and I apparently misunderstood some things.

Last edited:

- #5

matt grime

Science Advisor

Homework Helper

- 9,395

- 3

- #6

- 1,030

- 4

But I haven't read anything on the "norm" of surreal numbers; maybe it's defined differently.

- #7

matt grime

Science Advisor

Homework Helper

- 9,395

- 3

It does.If a metric needs to be real-valued,

and if the surreals form a proper class

they do - apparently.

Checking the definitions is always a good idea.

- #8

- 1,030

- 4

I guess that answers the question about complete ordered fields.

- #9

- 355

- 3

(Rehash of my earlier post... only this time, correct)

Ignoring that the surreals form a proper class rather than a set, you'd need to prove that you can't define a metric on them that turns them into a complete ordered field.

One way to see that the surreal numbers are not a complete ordered field is to use the axioms of a complete ordered field to show that they satisfy the archimedean property, which says that there are no infinitesimals other than 0 and no infinities. Since the surreal numbers have non-zero infinitesimals and infinities, they cannot be a complete ordered field.

Ignoring that the surreals form a proper class rather than a set, you'd need to prove that you can't define a metric on them that turns them into a complete ordered field.

One way to see that the surreal numbers are not a complete ordered field is to use the axioms of a complete ordered field to show that they satisfy the archimedean property, which says that there are no infinitesimals other than 0 and no infinities. Since the surreal numbers have non-zero infinitesimals and infinities, they cannot be a complete ordered field.

Last edited:

- #10

Hurkyl

Staff Emeritus

Science Advisor

Gold Member

- 14,916

- 19

(hint... can you find one bigger than everything in that set?)

- #11

- 1,030

- 4

- #12

Hurkyl

Staff Emeritus

Science Advisor

Gold Member

- 14,916

- 19

Or by trichotomy....

- #13

- 1,030

- 4

How does it violate trichotomy?

- #14

Hurkyl

Staff Emeritus

Science Advisor

Gold Member

- 14,916

- 19

You've asserted:

X is the set of all surreal numbers

{X | } is a surreal number

and it's a basic fact of surreal numbers that

{X | } is larger than every member of X

so....

Now that I think of it, I'm pretty sure it's also a theorem that every surreal number is a well-founded set.

X is the set of all surreal numbers

{X | } is a surreal number

and it's a basic fact of surreal numbers that

{X | } is larger than every member of X

so....

Now that I think of it, I'm pretty sure it's also a theorem that every surreal number is a well-founded set.

Last edited:

- Last Post

- Replies
- 3

- Views
- 1K

- Replies
- 0

- Views
- 2K